重载时未在C ++中的范围错误中声明

Question

尝试运行此代码时出现错误

In function 'int main()':
error: 'area' was not declared in this scope

我找不到解决该问题的明确方法。

#include <iostream>

using namespace std;

int main() {

    area(13.3, 67.4);

    return 0;
}

void area(int a, int b){
    cout << "The area is " << a * b << endl;
}

void area(float a, float b){
    cout << "The area is " << a * b << endl;
}

void area(double a, double b){
    cout << "The area is " << a * b << endl;
}

Answer 1

您需要先声明函数，然后才能使用它们。

要么向前声明它们：

#include <iostream>

using namespace std;

// forward declarations
void area(int a, int b);
void area(float a, float b);
void area(double a, double b);

int main() {
    area(13.3, 67.4);
    return 0;
}

void area(int a, int b){
    cout << "The area is " << a * b << endl;
}

void area(float a, float b){
    cout << "The area is " << a * b << endl;
}

void area(double a, double b){
    cout << "The area is " << a * b << endl;
}

否则，将实现移到main()上方：

#include <iostream>

using namespace std;

void area(int a, int b){
    cout << "The area is " << a * b << endl;
}

void area(float a, float b){
    cout << "The area is " << a * b << endl;
}

void area(double a, double b){
    cout << "The area is " << a * b << endl;
}

int main() {
    area(13.3, 67.4);
    return 0;
}

话虽这么说，由于实现完全相同，只是数据类型不同，因此请考虑使用单个模板化函数：

#include <iostream>

using namespace std;

template<typename T>
void area(T a, T b){
    cout << "The area is " << a * b << endl;
}

int main() {
    area<double>(13.3, 67.4);
    return 0;
}

Answer 2

将函数原型放在主要函数之上，您应该会很好。